Question: Consider the polar curve $r=e^{\theta+1}$ over the interval $\left( 0,2\pi \right)$. At which value of $\theta$ does the graph of $r$ have a vertical tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{5\pi}{4}$ only (Choice B) B $\dfrac{\pi}{4}$ or $\dfrac{5\pi}{4}$ (Choice C) C $\dfrac{3\pi}{4}$ only (Choice D) D $\dfrac{3\pi}{4}$ or $\dfrac{7\pi}{4}$
Answer: A vertical line has an undefined slope. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dx}{d\theta}=0$ but don't make $\dfrac{dy}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be indeterminate). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={e^{\theta+1}}\cos(\theta) \\\\ y&={e^{\theta+1}} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=e^{\theta+1}\sin(\theta) \\\\ \dfrac{dy}{d\theta}&=e^{\theta+1}(1)\sin(\theta)+e^{\theta+1}\cos(\theta) \\\\ &=e^{\theta+1}(\sin(\theta)+\cos(\theta)) \\\\ \\\\ x(\theta)&=e^{\theta+1}\cos(\theta) \\\\ \dfrac{dx}{d\theta}&=e^{\theta+1}(1)\cos(\theta)+e^{\theta+1}(-\sin(\theta)) \\\\ &=e^{\theta+1}(\cos(\theta)-\sin(\theta)) \end{aligned}$ Now let's solve $\dfrac{dx}{d\theta}=0$ on the interval $\left( 0,2\pi \right)$. $\begin{aligned} \dfrac{dx}{d\theta}&=0 \\\\ e^{\theta+1}(\cos(\theta)-\sin(\theta))&=0 \\\\ \cos(\theta)-\sin(\theta)&=0 \\\\ \cos(\theta)&=\sin(\theta) \end{aligned}$ Within our interval, our possible solutions are $\theta=\dfrac{\pi}{4}$ and $\theta=\dfrac{5\pi}{4}$. Finally, we evaluate $\dfrac{dx}{d\theta}$ for our two possible values of $\theta$ and require that $\dfrac{dx}{d\theta}\ne0$. $\begin{aligned} \left.\dfrac{dy}{d\theta} \right| _{{\theta =\dfrac{\pi }{4}}}&=e^{^{\left({\dfrac{\pi}{4}}\right)\normalsize +1}}\cdot\left(\sin\left({\dfrac{\pi}{4}}\right)+\cos\left({\dfrac{\pi}{4}}\right)\right) \\\\ &=e^{^{\left({\dfrac{\pi}{4}}\right)\normalsize +1}}\cdot\left(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\right) \\\\ &=\sqrt{2}e^{^{\left({\dfrac{\pi}{4}}\right)\normalsize +1}} \\\\ \\\\ \left.\dfrac{dy}{d\theta} \right| _{{\theta =\dfrac{5\pi }{4}}}&=e^{^{\left({\dfrac{5\pi}{4}}\normalsize +1\right)}}\cdot\left(\sin\left({\dfrac{5\pi}{4}}\right)+\cos\left({\dfrac{5\pi}{4}}\right)\right) \\\\ &=e^{^{\left({\dfrac{5\pi}{4}}\normalsize +1\right)}}\cdot\left(-\dfrac{\sqrt{2}}{2}+\left(-\dfrac{\sqrt{2}}{2}\right)\right) \\\\ &=-\sqrt{2}e^{^{\left({\dfrac{5\pi}{4}}\normalsize +1\right)}} \end{aligned}$ The graph of $r$ has vertical tangents at $\theta=\dfrac{\pi}{4}$ or $\theta=\dfrac{5\pi}{4}$. The graphs of the tangents are shown with the figure at different scales. ${10}$ ${20}$ ${30}$ ${40}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$ ${200}$ ${400}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$